Load securement pet Fmcsa requirement

16,000lb machine needs 50% WLL to secure it per 393.106(d) for a total of 8,000 WLL required.
0.435 X 8,000 = 3480 pulling forwards.
0.5 x 8,000 = 4,000 pulling backwards.
0.25 x 8,000 = 2,000 pulling to each side.

393.102 Section 2 is where you get those numbers from. IGNORE SECTION 1 as it only applies if you are basing your numbers off the breaking strength of the tiedowns.

Read all of 393.102 and you get the the end:

(c) Equivalent means of securement. The means of securing articles of cargo are considered to meet the performance requirements of this section if the cargo is:

(1) Immobilized, such so that it cannot shift or tip to the extent that the vehicle's stability or maneuverability is adversely affected; or

(2) Transported in a sided vehicle that has walls of adequate strength, such that each article of cargo within the vehicle is in contact with, or sufficiently close to a wall or other articles, so that it cannot shift or tip to the extent that the vehicle's stability or maneuverability is adversely affected; or

(3) Secured in accordance with the applicable requirements of §§ 393.104 through 393.136.

^^^^^^^^^^^^
The bold and underlined part above 393.102 (c) (3) means as long as you meet the rest of the securement rules you are also good here. You can effectively ignore everything else in the section before this line as long as you follow the rest of the applicable securement laws from 393.104 to 393.136.

 

Gotcha. I see my mistake.

 

The 0.8g and 0.5g is how they come up with the minimum aggregate WLL. AI, not knowing any better, converted to how much WLL is required in the braking situation. The publishing of 393.102, minimum performance criteria, explains how they came up with requiring loads to have a minimum aggregate WLL of 50% of the cargo weight, not any calculations you are required to do.